Capturing the replaced text using re.sub in python

Question

I am trying to create a report for all the text replacements done by the program using re.sub. I am not able to figure out how can I capture the replaced text in to a variable. Can any of you please help me in doing this? Please find the below code

import re

Report_file = open("report.txt", "w")
st = '''<item><AP>item1</AP><AP>Item2</AP><AP>item3</AP><AP>Item4</AP></item>'''

outval = re.sub(r'(?i)item1', "value1", st)
outval = re.sub(r'(?i)item2', "value2", outval)
outval = re.sub(r'(?i)item3', "value3", outval)  
print outval

I want the report file in the below format

OLD: item1
NEW: value1

OLD: item2
NEW: value2

OLD: item3
NEW: value3

Answer 1

You need to use a function instead of a replace pattern instead:

def build_replacer(replacement):
    def replace(match):
        print match.group(), replacement
        return replacement
    return replace

Then run:

outval = re.sub(r'(?i)item1', build_replacer("value1"), st)
outval = re.sub(r'(?i)item2', build_replacer("value2"), outval)
outval = re.sub(r'(?i)item3', build_replacer("value3"), outval)  

and it'll print the original text and it's replacement.

This then gives:

>>> st = '''<item><AP>item1</AP><AP>Item2</AP><AP>item3</AP><AP>Item4</AP></item>'''
>>> outval = re.sub(r'(?i)item1', build_replacer("value1"), st)
item1 value1
>>> outval = re.sub(r'(?i)item2', build_replacer("value2"), outval)
Item2 value2
>>> outval = re.sub(r'(?i)item3', build_replacer("value3"), outval)  
item3 value3
>>> outval
'<item><AP>value1</AP><AP>value2</AP><AP>value3</AP><AP>Item4</AP></item>'

Instead of printing you could also store that information elsewhere, of course.

The build_replacer() function just returns a new function, replace(), which is what re.sub() will use whenever it finds a match. Instead of directly replacing the matched text, it asks the function what to use as a replacement text.

The reason we use build_replacer() here as a nested function, is so we can store the fixed replacement text somewhere and re-use the same replacement function over and over again without having to hardcode the replacement text.

Answer 2

In your question and in the answer, you'll have to write as many instructions

outval = re.sub(r'(?i)item3', .......... )  

as there are items to replace.

What if there are 56 items to replace ?

.

In my following solution , there are 5 items to replace, but the instruction
r.sub(fruiting,text) is written only one time:

text = '''
OR 125
BA  48
Pr 12
ba 4
Cherry 147
Ba   10
Or 7
OR 6
Orange 2
AP 9
PR          3
Banana 101
or 22
pR 13
'''

. import re

the_items = ('OR','BA','AP','PR','CH')
new_items = ('Orange','Banana','Apple','Pear','Cherry')
corresp = dict(zip(the_items,new_items))

r = re.compile('(%s) *(\d+)' % '|'.join(the_items),
               re.IGNORECASE)

def fruiting(ma,longname = corresp):
    fresh = '%-12s %s' % (longname[ma.group(1).upper()],
                          ma.group(2) ) 
    tu = ('OLD: %r\n'
          'NEW: %r\n'
          %
          ( ma.group(),fresh) )
    print tu
    return fresh

print '%s%s' % (text, r.sub(fruiting,text))

result

OLD: 'OR 125'
NEW: 'Orange       125'

OLD: 'BA  48'
NEW: 'Banana       48'

OLD: 'Pr 12'
NEW: 'Pear         12'

OLD: 'ba 4'
NEW: 'Banana       4'

OLD: 'Ba   10'
NEW: 'Banana       10'

OLD: 'Or 7'
NEW: 'Orange       7'

OLD: 'OR 6'
NEW: 'Orange       6'

OLD: 'AP 9'
NEW: 'Apple        9'

OLD: 'PR          3'
NEW: 'Pear         3'

OLD: 'or 22'
NEW: 'Orange       22'

OLD: 'pR 13'
NEW: 'Pear         13'

.

OR 125
BA  48
Pr 12
ba 4
Cherry 147
Ba   10
Or 7
OR 6
Orange 2
AP 9
PR          3
Banana 101
or 22
pR 13

Orange       125
Banana       48
Pear         12
Banana       4
Cherry 147
Banana       10
Orange       7
Orange       6
Orange 2
Apple        9
Pear         3
Banana 101
Orange       22
Pear         13